4.8 Ratio and Proportions:
4.8 .1 Introduction:
We come across situations where we need to
distribute income/share profit in unequal proportions.
Sometimes, we may have to compare performance of
teams and individuals (For example a batsmen in a cricket match has scored more
than twice the score of another player)
In 6.4, we have seen that the centroid cuts the
median in unequal terms (two to one).
Under these circumstances study of the concept
‘ratio and proportion’ is useful.
The ratio of two to one is written as 2:1.
Similarly the ratio of a to b is written as a:b.
If a and b are two non zero quantities of same unit
and same kind, then the fraction a/b is called the ‘ratio’
of a to b and is written as a:b (we can also express
a/b=k).
a and b are called ‘terms’ of the ratio. Moreover the first term a is called ‘antecedent’
and the second term b is called ‘consequent’.
While comparing two quantities in
terms of ratio, we need to note the following:
1.
Ratio is always a pure number and does not have any unit of
measurement.
2.
Since ratios are pure numbers, they follow the normal
rules of addition/subtraction/multiplication/division of real numbers.
3.
Ratio is normally expressed in its lowest terms (10:40 is
expressed as 1:4)
4.
When we compare two quantities, they must be converted to
same unit of measurement before comparison
(For example if 1 hour 15 minutes and 45 minutes are to be compared, both of them need to be converted to
hours or minutes)
Properties of ratios:
1. If a:b then for m _{} 0, ma:mb and a/m = b/m
(_{} they all represent the same fraction a/b)
2. Given two ratios a:b =
c:d then ad = bc (_{} a/b = c/d)
3. Given two ratios a:b
> c:d then ad > bc (_{} a/b > c/d)
4. Given two ratios a:b
< c:d then ad < bc (_{} a/b < c/d)
Example: Can you solve the following problem?
The marks obtained by 4 students in an examination
are as follows:
Ram:Shyam = _{}:_{}
Shyam:Gopal = _{}: _{}
Gopal:Rama = _{} : _{}
Find the ratio of marks of Ram:Shyam:Gopal:Rama.
If Ram had got 42 marks find the marks of others.
Working:
Since each individual ratio do not have common
terms, we need to have common terms for comparison.
Ram:Shyam = _{}:_{} = 3/2:5/2 = 3:5 (_{}LCM of denominators of two ratios are 2)
= 21:35 (by multiplying both terms of the ratio by
7)
Shyam:Gopal = _{}: _{} = 7/4:16/5 =
35/20:64/20 = 35:64 (_{}LCM of denominators of two ratios are 20)
Gopal:Rama = _{} : _{} = 32/11: 43/22 =
64/22:43/22=64:43 (_{}LCM of denominators of two ratios are 22)
Thus
Ram : Shyam : Gopal : Rama=
21:35:64:43
Since 42 is Ram’s mark and 42 = 21*2 we multiply
each term of the ratio by 2
Hence
Ram : Shyam : Gopal : Rama=
21:35:64:43= 21*2:35*2:64*2:43*2
Thus their marks are 42, 70, 128
and 86 respectively.
4.8 Problem 1: Present ages of sister and brother is 14 years and 10
years respectively. After how many years will the ratio of their age becomes
5:4?
Solution:
Let x be the number of years after which the ratio
becomes 5:4.
After x years sister’s age
will be 14+x and brother’s age will be 10+x
It is given that the ratio after x years will be
5:4
_{} 14+x/10+x = 5/4
_{} 4(14+x) = 5(10+x) (By
cross multiplication)
_{}56+4x = 50+5x
_{} x = 6 (By
transposition)
Verification: After 6 years sister’s age will be 20 and brother’s age
will be 16 and 20:16 = 5:4 which is as given in the problem and hence the
solution is correct.
4.8 Problem 2: An ancient art piece of alloy weighing 10kg has 65%
copper. Another piece of alloy weighing 15kg contains 70% copper. The two pieces
are melted together and a new piece of alloy is cast. What is the % of copper
in the new piece thus cast?
Solution:
First we need to find the mass of copper in both
alloys.
Let copper be x kg in the first alloy
_{} x/10
= 65/100
_{} x =6.5kg
Let copper be y kg in the second alloy
_{} y/15
= 70/100
_{} y =10.5kg
Hence in the new piece of 25kg (10kg+15kg) mass of
copper is 17kg (6.5kg+10.5kg)
Therefore the ratio of mass of copper to alloy is
17:25
But 17/25 =68/100
Therefore the new piece contains 68% copper.
4.8 .2 Properties
of equal ratio:
If 4:5 = 8:10 we notice that the following are all true
No 

Reason 
1 
5:4 = 10:8 
5/4
=10/8 
2 
4:8 = 5:10 
4/8=5/10 
3 
(4+5)/5 =(8+10)/10 
9/5
=18/10 
4 
(45)5 =(810)/10 
1/5
= 2/10 
5 
(4+5)/(45) =(8+10)/(810) 
9/(1)
= 18/(2) 
Did you notice that the ratios can be
negative also?
In general, if a:b = c:d (a/b=c/d)
then the following are all true (m is any real number)
No 
Equivalent Ratio 
Property name 
Proof 
Which means or implies 

1 
b:a
= d:c 
Invertendo 
b/a
= d/c 
bc=ad 

2 
a:c
= b:d 
Alternendo 
a/c
=b/d 
ad=bc 

3 
(a+mb):b = (c+md):d 
Componendo 
(a/b)+m = (c/d)+m 
(a+mb)/b = (c+md)/d or (a+mb)/(c+md)
= b/d 

4 
(amb):b = (cmd):d 
Dividendo 
(a/b)m = (c/d)m 
(amb)/b = (cmd)/d or (amb)/(cmd) = b/d 

5 
(a+mb):(amb)
= (c+md):(cmd) 
Componendo
 Dividendo 
(a+mb)/(c+md)
= b/d =(amb)/(cmd)
From ratio in Sl No 3 and 4 

4.8 Problem 3: If (x+3y)/(x+y) =3/2 then find
x/y, x^{2}+y^{2}/ x^{2}y^{2}
, x^{3}+y^{3}/ x^{3}y^{3}
Solution:
(x+3y)/(x+y) =3/2 (given)
By Componendo – Dividendo,
(x+3y)+(x+y)/(x+3y)(x+y) = (3+2)/(32)
= 5/1
_{} (2x+4y)/2y = 5/1
_{} 2x+4y = 10y
_{} 2x = 6y
_{} x/y = 6/2 = 3/1
_{} x^{2}/y^{2}
= 9/1; x^{3}/y^{3}= 27/1
Since x^{2}/y^{2} = 9/1
_{} By Componendo –
Dividendo,
x^{2}+y^{2}/ x^{2}y^{2
} = (9+1)/(91) = 10/8 =5/4
Since x^{3}/y^{3}= 27/1
_{} By Componendo –
Dividendo,
X^{3}+y^{3}/ x^{3}y^{3}
= (27+1)/(271) = 28/26 =14/13
4.8 Problem 4: Solve the following equations:
(x^{2}16x+63)/ (x^{2}6x+8)
= (x^{2}16x+60)/ (x^{2}6x+5)
Solution:
By Alternendo
(x^{2}16x+63)/(x^{2}16x+60)
= (x^{2}6x+8)/(x^{2}6x+5)
By Dividendo
{(x^{2}16x+63)(x^{2}16x+60)}/ (x^{2}16x+60)
= {(x^{2}6x+8)(x^{2}6x+5)}/ (x^{2}6x+5)
_{} 3/(x^{2}16x+60)
= 3/(x^{2}6x+5)
_{}(x^{2}16x+60) = (x^{2}6x+5)
_{} 10x = 55 (By
transposition)
_{} x = 11/2
4.8 Problem 5: The work done by (x3) men in (2x+1) days and the
work done by (2x+1) men in (x+4) days are in the ratio of 3:10. Find the value
of x.
Solution:
A man day is defined as the unit of work done by
one person in one day.
Work done by (x3) men in (2x+1) days = [(x3)(2x+1)] man days
Work done by (2x+1) men in (x+4) days = [(2x+1)
(x+4)] man days
Assuming that the work done is same, we can say
that
[(x3)(2x+1)]/ [(2x+1)
(x+4)] = 3/10
Since (2x+1) cannot be zero (if it is 0 then the number
of men will be half which can not be true), we can cancel common factors in LHS
of the above statement.
[(x3)(2x+1)]/
[(2x+1)(x+4)]=
3/10
Thus
(x3)/(x+4) =3/10
10x30 = 3x+12
7x = 42
_{}x = 6
4.8 .3 Theorem on equal ratio:
If a/b = c/d = e/f = g/h
…….. and k, l, m, n, … are any numbers then
a/b = c/d = e/f = (ak+cl+em+gn…….)/(bk+dl+fm+hn….)
Proof:
Let a/b = c/d
_{} ad
= bc
Let us take the term a(bk+dl)
a(bk+dl)
=abk+adl (by Expansion)
=abk+bcl (_{} bc=ad)
=b(ak+cl) (Take the common factor ‘b’ out)
_{} a(bk+dl)
= b(ak+cl)
_{} a/b
= (ak+cl)/(bk+dl)
By extending this proof from two terms to more number of terms
we have
a/b = c/d = e/f =
(ak+cl+em+gn…….)/(bk+dl+fm+hn….)
4.8 Problem 6: Solve (12x^{2}20x+21)/ (4x^{2}+4x+15) =
(3x5)/(x+1)
Solution:
We can observe that LHS has terms in x^{2}
but RHS has terms only in x. Also we may note that if we multiply both the numerator
and the denominator of RHS by 4x
we get terms in x^{2}. Thus
it is logical to multiply both numerator and denominator of RHS by 4x.
However, we can only do this when x_{}0
Can x be zero?
If x = 0 then LHS = 21/15 and RHS = 5/1
Since LHS _{} RHS, x cannot be zero.
RHS = {(3x5)/(x+1)}*(4x/4x) = (12x^{2}20x)/(4x^{2}+4x)
_{} The given equation can
be rewritten as follows with each ratio being equal to k.
(12x^{2}20x+21)/(4x^{2}+4x+15)
= (12x^{2}20x)/(4x^{2}+4x) = k
By theorem on equal ratios
{(12x^{2}20x+21)  (12x^{2}20x)}
/ {(4x^{2}+4x+15) (4x^{2}+4x)} = k
_{} 21/15 = k=7/5
But k = (3x5)/(x+1)
_{} 7/5 = (3x5)/(x+1)
_{} 7x+7
= 15x25
_{}32=8x
_{}x=4
Verification: substitute x=4 in
the given equation
LHS = (19280+21)/(64+16+15)
= 133/95=7/5
RHS = (125)/(4+1) = 7/5
Since LHS=RHS our solution is correct.
4.8 Problem 7: If y/(b+ca) = z/(c+ab) =
x/(a+bc) then prove that a/(z+x) = b/(x+y) = c/(y+z)
Solution:
We notice that the given ratios have x, y and z in
numerator and a, b and c in denominator. But the ratios we need to prove have
a, b and c in numerator and x, y and z in denominator. Thus our working will be
easy if we do invertendo of the given ratios.
_{}By invertendo and making each ratio equal to k we have
(b+ca)/y = (c+ab)/z = (a+bc)/x = k
By applying the theorem on equal ratios on last two
terms we have
k = {(c+ab)+(a+bc)}/(z+x)
= 2a/(x+z)
Similarly we can prove k = 2b/(x+y) and k = 2c/(y+z)
_{} k = 2a/(x+z) =2b/(x+y)
= 2c/(y+z)
By dividing each of the above ratios by 2 we get
a/(x+z) = b/(x+y) = c/(y+z)
4.8 .3 Proportion:
If a/b = c/d then the numbers a, b, c and d are
said to be in proportion and we write them as a:b::c:d.
In such a relationship we call a and d as ‘extremes’
and b and c as middle terms or ‘means’
What if a/b = b/c (i.e. a:b
= b:c) such that
b^{2 }= ac and b = SQRT(ac)
In such case we say a, b and c are in ‘continued proportion’ and b is called ‘geometric mean’ or ‘mean proportional’ between
‘a and c’
4.8 Problem 8: The middle of
three numbers in continued proportion is 24 and the sum of first and the third
is 52. Find the numbers.
Hint:
Let a and c be the first
and the third number. Thus the three numbers in continued proportion are a, 24
and c.
_{} ac
= 24^{2}=576
It is also given that a+c = 52
_{} c =52a
By substituting this value in ac we get the
following equation to solve
a^{2}52a+576 = 0
After factorisation we find that a=16 and a=36 are
the roots of the equation and consequently 36 or 16 are the values of c
respectively.
Hence (16,24,36) or
(36,24,16) are the three required numbers.
4.8 .4 kmethod:
In this method we equate each of the ratio to a number k and then solve the problem. Hence this
method is called as kmethod.
4.8 Problem 9: Five numbers are
in continued proportion. The product of the 1^{st} and the 5^{th}
of them is 324 and the sum of the 2^{nd} and
the 4^{th} is 60. Find the numbers.
Hint:
Let a, b, c, d and e be
the numbers.
It is given that ae=324 and b+d=60
Let the ratio of the numbers be k
If we can find the value of k we could also find
the value of the numbers
a/b=b/c=c/d=d/e=k
d=ek, c=dk=ek^{2}, b=ck=ek^{3} and a=bk=ek^{4}
Let us take a=ek^{4}
_{}(324/e) = ek^{4 }(_{}ae = 324)
_{}324 = e^{2}k^{4 }(_{}324=18*18)
_{}e =18/k^{2}
ek^{3}+ek = 60 (_{}b+d=60)
_{}(18/k^{2})k^{3}+(18/k^{2})k=60
(_{}e=18/k^{2})
_{}18k + 18/k = 60
_{}3k+3/k=10
_{} 3k^{2}10k+3=0
_{} k = 3 and k = 1/3 are the
roots of the above equation
If k = 3 then we have e=2,
a=162, b=54, c=18 and d=6
If k =1/3 then we have e=162,
a=2, b=6, c=18 and d=54
_{} The required numbers
are (162,54,18,6,2) or (2,6,18,54,162).
4.8 Problem 10: If a:b::c:d show that
1. (a+b):(c+d) = SQRT(a^{2}+b^{2}):
SQRT(c^{2}+d^{2})
2. (a^{2}+c^{2}):(ab+cd)
= (ab+cd):(b^{2}+d^{2})
Solution:
Let a/b = c/d = k
_{} a = bk,
c = dk
Part 1:
LHS = (bk+b)/(dk+d) = b/d
RHS = SQRT(b^{2}k^{2}+b^{2})/SQRT(d^{2}k^{2}+d^{2})
= b.SQRT(k^{2}+1)/
d.SQRT(k^{2}+1)= b/d
Thus LHS=RHS
Part 2:
LHS = (b^{2}k^{2} + d^{2}k^{2})/(b^{2}k + d^{2}k) = k^{2}/k = k
RHS = (b^{2}k + d^{2}k)/(b^{2 }+ d^{2})
= k
Thus LHS=RHS
4.8 Summary of learning
If
a/b = c/d = e/f = g/h . . . . and k, l, m, n… are any numbers then 

No 
Points to remember 
1 
b:a
= d:c 
2 
a:c
= b:d 
3 
(a+mb):b
= (c+md):d 
4 
(amb):b
= (cmd):d 
5 
(a+mb):(amb)
= (c+md):(cmd) 
6 
a:b = c:d = e:f = (ak+cl+em+gn..):(bk+dl+fm+hn.
.) 